// 给你一个由 '1'（陆地）和 '0'（水）组成的的二维网格，请你计算网格中岛屿的数量。
//
// 岛屿总是被水包围，并且每座岛屿只能由水平方向和/或竖直方向上相邻的陆地连接形成。
//
// 此外，你可以假设该网格的四条边均被水包围。
//
//
//
// 示例 1：
//
//
// 输入：grid = [
//  ["1","1","1","1","0"],
//  ["1","1","0","1","0"],
//  ["1","1","0","0","0"],
//  ["0","0","0","0","0"]
// ]
// 输出：1
//
//
// 示例 2：
//
//
// 输入：grid = [
//  ["1","1","0","0","0"],
//  ["1","1","0","0","0"],
//  ["0","0","1","0","0"],
//  ["0","0","0","1","1"]
// ]
// 输出：3
//
//
//
//
// 提示：
//
//
// m == grid.length
// n == grid[i].length
// 1 <= m, n <= 300
// grid[i][j] 的值为 '0' 或 '1'
//
// Related Topics 深度优先搜索 广度优先搜索 并查集 数组 矩阵 👍 1924 👎 0

package leetcode.editor.cn;
// Java：岛屿数量
public class P200NumberOfIslands {
    public static void main(String[] args) {
        Solution solution = new P200NumberOfIslands().new Solution();
        // TO TEST
    }
    // leetcode submit region begin(Prohibit modification and deletion)
    class Solution {
        public int numIslands(char[][] grid) {
            if (grid == null || grid.length == 0) {
                return 0;
            }

            int nr = grid.length;
            int nc = grid[0].length;
            int num_islands = 0;
            for (int r = 0; r < nr; ++r) {
                for (int c = 0; c < nc; ++c) {
                    if (grid[r][c] == '1') {
                        ++num_islands;
                        this.dfs(grid, r, c);
                    }
                }
            }

            return num_islands;
        }

        void dfs(char[][] grid, int r, int c) {
            int nr = grid.length;
            int nc = grid[0].length;

            if (r < 0 || c < 0 || r >= nr || c >= nc || grid[r][c] == '0') {
                return;
            }

            grid[r][c] = '0';
            this.dfs(grid, r - 1, c);
            this.dfs(grid, r + 1, c);
            this.dfs(grid, r, c - 1);
            this.dfs(grid, r, c + 1);
        }
    }
    // leetcode submit region end(Prohibit modification and deletion)

}
